Use the Gauss-Jordan method to solve the system of equations.
-$$
\begin{array}{l}
x+y+z=7 \
x-y+2 z=7 \
2 x+3 z=14
\end{array}
$$
A) $ \left(\frac{-3 z+14}{2}, \frac{z}{2}, z\right) $
B) $ \left(\frac{-3 z-14}{2}, \frac{z}{2}, z\right) $
C) $ \left(\frac{-3 z+14}{2}, 2 z, z\right) $
D) $ \left(\frac{3 z+14}{2}, \frac{z}{2}, z\right) $

Question

Quizplus · Accepted Answer

To solve the system using the Gauss-Jordan method, we first write the augmented matrix and then apply row operations to get it in reduced row echelon form. The system of equations is:$$ \begin{align*}x + y + z &= 7 \x - y + 2z &= 7 \2x + 0y + 3z &= 14\end{align*} $$The augmented matrix is:$$ \left[ \begin{array}{ccc|c}1 & 1 & 1 & 7 \1 & -1 & 2 & 7 \2 & 0 & 3 & 14\end{array} \right] $$Applying row operations to simplify:1. Subtract row 1 from row 2.2. Subtract 2 times row 1 from row 3.3. Further simplify to isolate each variable in a row.After simplification, we find that the system does not have a unique solution for $x$, $y$, and $z$, but rather expresses $x$ and $y$ in terms of $z$. The correct solution is in the form where $x$ and $y$ are expressed in terms of $z$, which matches option A: $ \left(\frac{-3 z+14}{2}, \frac{z}{2}, z\right) $. This indicates that for any value of $z$, we can find corresponding values of $x$ and $y$ that satisfy the system of equations.

Use the Gauss-Jordan Method to Solve the System of Equations x+y+z=7x−y+2z=72x+3z=14\begin{array}{l}x+y+z=7 \\x-y+2 z=7 \\2 x+3 z=14\end{array}x+y+z=7x−y+2z=72x+3z=14​

Use the Gauss-Jordan Method to Solve the System of Equations $\begin{array}{l}x+y+z=7 \\x-y+2 z=7 \\2 x+3 z=14\end{array}$